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Dahasolnce [82]
3 years ago
5

Consider the following unbalanced particulate representation of a chemical equation:

Chemistry
1 answer:
nevsk [136]3 years ago
3 0

Answer:

H2 + I2 --> 2HI

Explanation:

The two reactants are diatomic molecules because they contain two atoms of the same element. Therefore, they would need to have a subscript of "2" next to their symbols.

When balancing an equation, you want the same amount and type of atoms on both sides. By adding a coefficient of "2" in front of the product, two H's and two I's are now on both sides.

You might be interested in
Define not matter and give examples​
VLD [36.1K]

Answer:

Everything that has mass and takes up space is matter. Every day, you encounter phenomena that either don't have mass or don't take up space. They are non-matter. Basically, any type of energy or any abstract concept is an example of something that is not matter.

An apple.

A person.

A table.

Air.

Water.

A computer.

Paper.

Iron.

hope this helped you

3 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
Bond [772]

Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
3 years ago
URGENT!!! A solution in which there is very little solute dissolved in a solvent.
ella [17]

Answer:

Is called a diluted solution

Explanation:

Having little solute makes the dissolving process in a large amount of solvent very easy to mix therefore it dilutes in the solvent

4 0
3 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
What is the hydroxide-ion concentration in a solution formed by combining 200. ml of 0.16 m hcl with 300. ml of 0.091 m naoh at
iren2701 [21]
34 degrees long that way you pass the exam
8 0
3 years ago
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